# WKB and asymptotic behavior of second order differential equation

I want to study the large $x$ solution to a Riccati equation. After listening to the lectures on Mathematical Physics by Carl Bender, I have fallen in love with asymptotic analysis. But, by no means do I have a full understanding of it.

I transformed the Riccati equation into the following second order differential equation:

$$u''(x)-Q(x) u(x)=0\qquad Q(x)=\frac{n(n+2)}{4x^2}+\frac{a^2 b^2\,e^{-2x}}{x^{2n+1}},\enspace\text{with}\,\, a,b>0, \,\,n=\{0,1,2,\ldots\}$$

defined on $0<x<\infty$, subject to the initial condition: $u(1)=1$, $u'(1)=1$. The solution to this differential equation gives the solution y(x) to the Riccati equation: $$y_\text{sol}(x)=-\frac{(n+2)}{2a}x^{n+1} + \frac{x^{n+2}}{a}\frac{d}{dx}\ln u(x)$$

For concreteness take $n=0$ and $a=100$, $b=1$.

My Attempt

My numerical investigations revealed that the large $x$ behavior of $u$ is very insensitive to the initial conditions. (The hunch I got from this is that the dependence on the initial condition would appear in the negligible part of the solution). And $y_\text{sol}$ at large $x$ tends to a non-zero constant which I am interested in.

I apply WKB. But, before doing so I recognized this problem as arising from a 3-dimensional QM problem where $x$ is a radial coordinate. I recalled some obscure claim by Langer (1937) that in order to get an 'accurate WKB result' we must rescale $x$ by putting $x=e^z$ to make the domain of the problem over the entire real numbers $-\infty<z<\infty$. [I have no understanding of why this must be done from an asymptotic analysis point of view; can someone help me with this?]. Then if I put $u(x)=e^{z/2}w(z)$, I get a new second order differential equation

$$w''(z) - P(z)w(z)=0\qquad P(z)=\frac{(n+1)^2}{4}+e^{2z}\frac{a^2b^2\,e^{-2(e^{z})}}{(e^{z})^{2n+1}}$$

for which I can apply WKB. The end result (after transforming back to $u$ in $x$-space) is:

$$u(x)\sim A \frac{1}{Q_L(x)^{1/4}}\Big(e^{\int_1^x dx' \sqrt{Q_L(x')}}+B e^{-\int_1^x dx' \sqrt{Q_L(x')}}\Big)$$ Here, $Q_L = \displaystyle\frac{(n+1)^2}{4x^2}+\frac{a^2 b^2\,e^{-2x}}{x^{2n+1}}$ is slightly different from $Q$ due to the brief transformation to $z$-space, and $A$ and $B$ are constants fixed by boundary conditions.

Problems

When I plot this result against the numerically integrated solution for $u(x)$ and also for $y(x)$, I find that WKB does a fantastic job approximating the small $x$ behavior of the exact solution, but doesn't get quite get the large $x$ behavior correctly. WKB does get leading the asymptotic behavior: $u'(x)/u(x) \sim \frac{n+2}{2}\frac{1}{x}$ as $x\rightarrow\infty$. But I need the next-to-leading asymptotic form which I expect to be $\sim 1/x^{n+2}$. But WKB approximation tells me that it is $\sim e^{-2x}$, which is very subdominant compared to what I am expecting. I am puzzled by this because I thought WKB was supposed to uniformly get the right answer for all $x$! I tried even tried taking the WKB approximation to next to leading order in hopes of getting the expected next-to-leading order asymptotic behavior, but I still keep getting highly subdominant terms... what's going on??

Perhaps I should follow a different strategy to determine the asymptotic behavior?

Plot Here is a plot of $y(x)$ satisfying $y(1)=1$ for $n=0$ and $a=100$, $b=1$:

Magenta curve: numerically integrated. Blue curve: WKB result.

• Is the original Riccati equation $y'(x) + y(x)^2 - Q(x) = 0$? May I assume the initial condition is $y(1) = 1$? – Antonio Vargas Apr 1 '14 at 15:00
• The original Riccati equation is $y'(x) = -\frac{a}{x^{n+2}} y^2(x) + a b^2 \frac{e^{-2x}}{x^{n-1}}$. And there is no harm in choosing $y(1) = 1$ as the initial condition (that was the condition I used to generate the plot). – QuantumDot Apr 1 '14 at 17:06
• I'm only mildly familiar with the WKB method but I don't think it's used for large $x$ asymptotics like this. It's usually used to find asymptotic solutions over a finite spatial interval as some other parameter is sent to $\infty$, or at least that's where I've seen it proved that the asymptotic estimates are uniform with respect to $x$. – Antonio Vargas Apr 1 '14 at 18:43
• As for your problem I'm having a tough time getting anywhere. I can show that $$y(x) > \frac{(n+1)x^{n+1}}{(n+1+a)x^{n+1} - a} > \frac{n+1}{n+1+a}$$ for all $x > 1$, and I can find a much uglier upper bound, but the large $x$ behavior still escapes me. I'm interested to see an answer to the question as well. – Antonio Vargas Apr 1 '14 at 18:45